Chemistry 235 - Final Examinations Summer 2006 – answers
A GENERAL CHEMISTRY
answers are given from 1-20:
OH
Br
NHCH3
no reaction
O
H
Br
COOCH3
OH
N
COOH
N
H
Ph
COOH
Br
MeOH
COOH
O
Me
OH
COOH
HO
Cl
O
O
CH3NH
O
OH
O
EtO
Br
NH2
O
O
NH 2
OMe
OH
OH
O
NH2
B. REACTIONS AND REAGENTS [32 MARKS]
2. A single substance can often be converted into more than one closely related synthetic intermediate
by appropriate selection of experimental conditions. Provide experimental conditions to carry out
each of the indicated conversions [the answer requires the starting material to be converted
independently to each of the products]. Some of the conversions may require more than one step.
OH
Br
OH
Br
CH3CHO
Mg
Et2O
MgBr
O
H3O
OH
+
H3O+
OH
CH2Br
and:
CH3
CH3
Br
Br
NBS
Br2
hν
Fe
Br
3. Answer the following questions about the Diels Alder reaction. Remember, stereochemistry is
important. [8 marks]
a) Each of the following pairs of starting materials provides a single major Diels Alder product.
Give the formula of each one.
COOCH3
+
H3COOC
COOCH3
COOCH3
+
CHO
H
CHO
b) Give the starting materials which would yield the Diels Alder products shown here:
O
O H
O H
O
CN
NC
CO2CH3
COOMe
4. Provide a structure for each of the lettered intermediates [A - D] in the reaction sequence shown
below. [8 marks]
O
O
HO
OH
OEt
O
O
OEt
+
H
selective for ketone;
no reaction with ester
O
O
O
CH3MgI
H
selective reduction
ester to aldehyde
NaBH4
CH3OH
O
dry Et2O
B
H3O+
OH
A
H3O+
DIBAL-H
(one mole)
H2O
toluene
O
OH
remember - the hydrolysis
cleaves off the ketal group
OH
D
C
5. Consider the compound shown here:
O
O
a) Draw the structure of all possible enols of this compound
b) Give the product of its reaction with excess D2O and a catalytic amount of base (-OH)
c) Give the product of its reaction with one mole of Br2 and a catalytic amount of acid (H+)
d) Give the product which results when this compound is treated with sodium ethoxide (NaOEt)
d) Give the product of the crossed aldol reaction which takes place when the following two
compounds are reacted (for example with NaOMe/MeOH):
O
O
+
a)
OH
O
OH
b)
O
O
O
D
D
c)
O
O
Br
d)
e)
D
D
D DD D
D
D
O
O
6. There are fatal flaws in each step in the following sequence of reactions. Clearly identify what
is wrong with each reaction.
O
SO3H
SO3H
Cl
H2SO3
SO3H
H3O
NaBH4
AlCl3
+
EtOH
O
SO3H
Br
Br2
Mg
Fe
H3O+
SO3H
O
MeLi
H
HO
OH
dry Et2O
KMnO4
HO
heat
HCHO
OH
H2O
O
H3O+
dry Et2O
step 1: must be H2SO4
step 2: Friedel Crafts reactions cannot be carried out when only meta- substituents are present
step 3: NaBH4 does not reduce ketones to methylenes but to alcohol
step 4: reaction will not occur between the two substituents
step 5: cannot perform a Grignard reaction in the presence of the sulfonic acid
step 6: SO3H is removed by dilute acid, not dilute base
step 7: oxidation would cause both substituents to go to acid
step 8: reaction of organolithium with aldehyde does not yield ketone, but alcohol
C. MECHANISMS AND CONCEPTS [32MARKS]
DO FOUR OF QUESTIONS 7-12
7. Aromaticity
Use the Hückel rule to estimate whether each of the following compounds will exhibit
aromaticity or not. Give your reasons. The correct answer for the wrong reason will not get any
marks.
O
O
asssume it can be planar
anion is 2 electrons
total 10; n=2
AROMATIC
3
sp centres
each triple bond is 2
not conjugatged total 18; n=4
NOT AROMATIC AROMATIC
each oxygen is 2
total 10; n=2
AROMATIC
8. a) Provide an explanation for the following observations: p-N,N-dimethylaminotoluene
[below] undergoes nitration in mildly acidic solution to give 4-dimethylamino-3-nitrotoluene and
in strongly acidic solution to give 4-dimethylamino-2-nitrotoluene.
Me
Me
Me
N
Me
Me
N
NO2
Me
N
mild H+
strong H+
Me
NO2
Me
Me
In mild acid, no protonation of the nitrogen occurs, and thus the Me2N substituent acts normally:
Me
Me
Me
Me
δ+N
H
N
Me
Me
N
NO2
NO2
Me
Me
Me
The nitrogen is able to stabilize the ortho- or para- intermediate in the usual fashion, leading to the
normal substitution pattern.
In strong acid solution, the nitrogen protonates, and becomes a very deactivating, metadirecting group:
Me
Me
N
Me
Me
N
Me
H
Me
N
Me
H
Me
N
NO2
NO2
Me
Me
Me
Me
Both the methyl and the ammonium now direct to the other position.
b) 3-Methoxy-1,2,3-trimethylcyclopropene [shown below] has been found to react with
tetrafluoroboric acid to yield methanol and a compound C6H9+BF4-. What is the structure of this
product? Why is it formed so easily?.
CH3O
H3C
CH3
HBF4
C6H9
BF4
+
CH3OH
CH3
Ionization is particularly easy since the product is aromatic; (4n+2) where n=0
HBF4
CH3O
CH3
H
CH3O
BF4
-
CH3
CH3
-CH3OH
CH3
BF4
H3C
CH3
H3C
CH3
H3C
CH3
-
CH3
H3C
9. Rank the following compounds in order of ease of addition of a nucleophile [for example
methanol]; clearly indicating for which the reaction is easiest [fastest], and give a brief explanation
as to your choice of order.
F3 C
3
O
O
O
O
H
1 (best)
H
4 (worst)
2
Reaction depends upon the relative positive charge located on the carbonyl carbon, and upon steric
influences. Considering both of these criteria, aldehydes normally react faster than ketones
– the alkyl group is inductively electron donating, decreasing the *+ charge on carbonyl carbon,
making attack slower
– replacing a H (small) with an alkyl (larger) increases the steric problems in both addition and
product crowding, making aldehydes faster reacting.
Between the two aldehydes, CF3 will react much faster than CH3. Since F is strongly electron
withdrawing, *+ will be very large and addition will occur very easily.
Between the two ketones, we must compare isopropyl (2-propyl) and phenyl as substituents. Both
are large, but phenyl has the added feature of being able to delocalize the + charge through resonance
with the ring:
This makes it much less *+ on the carbon, and reaction the slowest of these four.
10. The acidity of carboxylic acids is dependent upon the nature and position of substituents. The pKa
values of the four acids given below each differ from the other by one pK unit or more. Rank the
four acids in order of acidity, from most acidic to least, and give a brief explanation for this
ranking. The actual pK values are 1.7, 2.9, 4.1, and 5.0.
Cl
O
O
O
OH
OH
OH
O
O2 N
OH
Cl
3
4
2
1 (most acidic)
Acidity depends upon the relative stability of the carboxylate ions. The substituents here can only
affect this stability through inductive effects; since the charge on the carboxylate is negative, electron
withdrawing groups will help stabilize the anion and electron donors will not. NO2 is a strong
electron withdrawing group, and as such will stabilize the anion most effectively and make the
corresponding acid the most acidic (1.7). Cl is a weaker electron withdrawer, but will still stabilize
and increase acidity; 2-Cl will be more acidic (2.9) than 3-Cl (4.1) since the inductive effect of a
substituent decreases rapidly with distance. Methyls are electron donors, and as such increases
electron density on the carboxylate anion, and thus this acid is the least acidic (5.0).
11. Give the mechanism of either of the reactions shown here. Remember to include all intermediate
structures and use proper arrows. [8 marks]
O
OMe
CH3OH
OH
or
O
H+
O
O
N
CH3Li
Et2O
H3O+
O
H+
H+
OH
O
H3C OH
CH3
OH
CH3
OH
H3C OH
2
H3C HOCH3
H3C OCH
3
H3C OCH
3
O
O
O
O
+
N Li
O
H+
H3O+
NH
H3O+
O
NH2
OH2
CH3Li
H+ NH
2
NH2
HO
O
H
O
N
O
O
O
H+
B
OH
O
O
H3C OH
NH3
OH
H OH2
O
O H
OH2
12 2,4-Dinitrochlorobenzene undergoes an easy nucleophilic aromatic substitution (as shown below)
to yield the methoxy compound. 3,5-Dinitrochlorobenzene (or, for that matter, chlorobenzene
itself) does not react under the same conditions. Using proper mechanistic diagrams, clearly
explain why this is so.
Cl
O2 N
OMe
NaOMe
O2 N
MeOH
NO2
NO2
Nucleophilic aromatic substitution occurs by an addition-elimination mechanism. In the initial
step, the nucleophile, methoxide ion, attacks the benzene ring – adding to it and forming an
intermediate which is negatively charged:
Cl
OMe
O2 N
O2 N
Cl
NO2
OMe
O 2N
Cl
OMe
O 2N
Cl
NO2
NO2
OMe
NO2
In the absence of substituents (in chlorobenzene itself), this anion is not sufficiently stable, even
with resonance delocalization, for the process to occur. Strong electron-withdrawing substituents must
be present at appropriate positions – i.e. where they can further delocalize the charge. This happens
readily for the compound above:
O2 N
Cl
OMe
O2 N
Cl
OMe
O
and
O
N
O
O
N
N
Cl
N
O
Cl
OMe
O
NO2
O
O
OMe
NO2
Each of the nitro groups can become involved and the anionic intermediate is greatly stabilized
– to the point where addition can occur. Note that if the position of the nitros is incorrect (as in the
3,5- isomer), no reaction occurs.
The process then completes by loss of the leaving group:
O2 N
Cl
OMe
OMe
O 2N
+
NO2
Cl
NO2
D SYNTHESIS
13. Complete THREE of the four syntheses shown below, using only the indicated starting materials
and any other necessary reagents, solvents, etc., remembering that you must make any organometallic
or Wittig reagents you with to use [this refers to RMgX or RLi, not to LiAlH4 ]. Each synthesis may
be carried out in five or less steps.
a)
from any organic starting
materials of 4 carbons or less
PBr3
OH
Li
Br
OH
H3O+
HCHO
Et2O
PBr3
OH
PPh3
Br
n-BuLi
PPh3
THF
PCC
OH
PPh3
O
CH2Cl2
THF
b) HO
O
O
O
or
O
O
OEt
or
O
EtO
OEt
and any other necessary reagents
O
O
NaOEt
OEt
BuBr
O
O
EtOH
NaOEt
O
HO
H3O+
MeLi
∆
and
dry Et2O
MeI + 2 Li
MeLi
dry Et2O
H3
O+
O
O
OEt
EtOH
OEt
MeI
COOH
c)
from benzene and
any necessary reagents
O
SO3H
SO3
KOH
H2SO4
∆
H2SO4
SO3H
OH
Cl2
Cl
OH
OH
Cl
Fe
OH
dil
H2SO4
Cl
Cl
∆
SO3H
d)
Cl
OH
from benzene and any
necessary reagents
Cl
Br
CH3COCl
Br2
AlCl3
Fe
O
O
Br
Mg
O
O
dry Et2O
CO2
HO
H3O+
COOH
∆
O
OH
BF3